O is the centre of a circle and AB is a tangent to the circle touching it at point B. If OB = 3 cm and OA = 5 cm, where A is an external point lying on the tangent line, then what is the length of segment AB (in cm)?

Introduction / Context:
This question tests your understanding of the geometry of a circle with a tangent line. A very important property in circle geometry is that the radius drawn to the point of tangency is perpendicular to the tangent. When a radius, a tangent segment, and the segment joining the centre to an external point are considered together, they naturally form a right triangle. This allows us to apply the Pythagoras theorem to find unknown lengths. Such problems are very common in aptitude tests and school level geometry because they combine fundamental circle properties with basic right triangle calculations.

Given Data / Assumptions:
• O is the centre of the circle.
• AB is a tangent to the circle at point B.
• OB is a radius of the circle, and OB = 3 cm.
• OA is the segment from the centre to the external point A on the tangent, and OA = 5 cm.
• The angle between OB and AB at B is 90 degrees because a radius is perpendicular to the tangent at the point of contact.
• We are required to find the length of AB in centimetres.

Concept / Approach:
Since OB is perpendicular to AB at B, triangle OAB is a right angled triangle with right angle at B. In that triangle, OA is the hypotenuse, OB is one leg, and AB is the other leg. By the Pythagoras theorem for a right triangle, (hypotenuse)^2 = (leg1)^2 + (leg2)^2. Therefore, OA^2 = OB^2 + AB^2. Rearranging this equation allows us to solve for AB^2 and then take the square root to get AB. This is a direct and standard application of both circle properties and the Pythagoras theorem.

Step-by-Step Solution:
Step 1: Recognise that OB ⟂ AB because a radius is perpendicular to the tangent at the point of tangency. Step 2: Conclude that triangle OAB is a right triangle with right angle at B. Step 3: Apply the Pythagoras theorem: OA^2 = OB^2 + AB^2. Step 4: Substitute OA = 5 cm and OB = 3 cm into the equation: 5^2 = 3^2 + AB^2. Step 5: Compute the squares: 25 = 9 + AB^2. Step 6: Rearrange to find AB^2: AB^2 = 25 - 9 = 16, so AB = √16 = 4 cm.

Verification / Alternative check:
We can confirm the result by quickly checking Pythagoras in reverse: OB^2 + AB^2 = 3^2 + 4^2 = 9 + 16 = 25, which is exactly OA^2, and OA is 5 cm, so 25 is correct. This matches the standard 3, 4, 5 Pythagorean triple, which further supports that our triangle dimensions are consistent. There is no contradiction with the circle and tangent properties, so the answer is reliable.

Why Other Options Are Wrong:
Option √34 cm would correspond to AB^2 = 34, which would give OA^2 = 9 + 34 = 43, not 25, so it contradicts OA = 5 cm. Option 2 cm would give OA^2 = 3^2 + 2^2 = 9 + 4 = 13, which again does not match 25. Option 8 cm would give OA^2 = 9 + 64 = 73, far from 25. Hence all of these violate the Pythagoras relation with the given OA and OB values.

Common Pitfalls:
A typical mistake is to forget that the right angle is at B and to misapply Pythagoras with the wrong segment as the hypotenuse. Another error is to use subtraction incorrectly, such as computing 3^2 - 5^2 or reversing the hypotenuse and a leg. It is crucial to correctly identify OA as the longest side since it connects the centre to an external point and thus must be larger than the radius. Remembering the radius tangent perpendicular property and carefully writing the Pythagoras equation helps avoid these issues.

Final Answer:
The length of segment AB is 4 cm.